Question

The equation which represents Freundlich adsorption isotherm is (x = amount of gas, m = mass of solid)

• A. \( \log\frac{x}{m} = \log p + \frac{1}{n}\log k \)
• B. \( \log\frac{x}{m} = \log k + \frac{1}{n}\log p \)
• C. \( \frac{x}{m} = k + \frac{1}{n}\log p \)
• D. \( \frac{x}{m} = \log p + \frac{1}{n}\log k \)

Hint:

- Freundlich Adsorption Isotherm (empirical): \( \frac{x}{m} = k p^{1/n} \). - \(x/m\): amount of gas adsorbed per unit mass of adsorbent. - \(p\): equilibrium pressure of adsorbate. - \(k, n\): constants (n>1). - Logarithmic form: \( \log\left(\frac{x}{m}\right) = \log k + \frac{1}{n}\log p \). This is linear: \( Y = C + MX \), where \( Y=\log(x/m) \), \( X=\log p \), slope \( M=1/n \), intercept \( C=\log k \).

Answer 1

The Correct Option is B

The Freundlich adsorption isotherm gives an empirical relationship between the amount of gas adsorbed by a unit mass of solid adsorbent (\(x/m\)) and the pressure (\(p\)) of the gas at a particular constant temperature.
The equation is: \[ \frac{x}{m} = k p^{1/n} \] where: - \(x\) is the mass of the gas adsorbed.
- \(m\) is the mass of the adsorbent.
- \(p\) is the equilibrium pressure of the gas.
- \(k\) and \(n\) are constants that depend on the nature of the adsorbent and the gas at a particular temperature.
\(n\) is typically greater than 1.
To get the logarithmic form, take the logarithm (base 10 or natural logarithm) of both sides: \[ \log\left(\frac{x}{m}\right) = \log(k p^{1/n}) \] Using properties of logarithms \( \log(AB) = \log A + \log B \) and \( \log(A^B) = B\log A \): \[ \log\left(\frac{x}{m}\right) = \log k + \log(p^{1/n}) \] \[ \log\left(\frac{x}{m}\right) = \log k + \frac{1}{n}\log p \] This equation is of the form \( y = c + mx' \), representing a straight line if \( \log(x/m) \) is plotted against \( \log p \), with slope \( 1/n \) and y-intercept \( \log k \).
This matches option (2).